Geometric meaning of parallel transport The definition of parallel transport of a vector $v^b$ along a curve $C$ with tangent field $\it{t}^a$ is given by Wald's GR as $$t^a \nabla_a v^b = 0$$ 
Is it correct to think of $\nabla_a v^b$ as orthogonal to the curve $C$, such that the "inner product" with $t^a (\nabla_a v^b)$ is zero? Is it analogous (or exactly the same) as the fact that the gradient of a curve is orthogonal to the tangent vector along the curve?
 A: Both $t^a$ and $v^b$ are vector fields defined (at least) on $C$ and $t^a$ points along $C$.
If $t^a\nabla_av^b=0$ everyewhere on $C$ and you take $v^b$ at a particular point $p$ on $C$ then $v^b$ at all other points on $C$ is given by parallel transporting that initial value of value of $v^b$ from $p$ along $C$ (ie along the tangent $t^a$).
Another way of thinking about it is that the covariant derivative in a given direction is the rate of change of the vector field wrt the parallel transported vector. Therefore if the vector field is not changing wrt parallel transported vector (ie $t^a\nabla_av^b=0$) then the field at each point is equal to the transported vector.
Penrose has a nice visual explanation of the relationship between parallel transport and the covariant derivative in Road to Reality Ch 14, p298:

A: To address the first question you can consider $$ \nabla_av^b := v^b ,_a$$ since $$ \nabla : \Gamma(E)  \rightarrow  \Gamma(E\otimes T^*M)$$
where E is any section(e.g. the tangent bundle in question)
and contracting it with the tangent vector field of the curve you get $$t^av^b,_a = 0 $$ this is similar to contracting a vector field with a dual vector
$$ t^av_a$$
So, yes, you can somewhat considered it as the "inner-product" since the contraction in the case of a vector and a dual vector is basically the familiar inner product.
About the second question. I'm not sure you can compare the gradient and the tangent like that. This might help clarify. 
https://math.stackexchange.com/questions/290903/difference-between-a-gradient-and-tangent
A: I think the answer to this is basically yes. You do have to be careful, because in general, you can't interpret inner products in relativity as measures of whether something is "orthogonal" to something else in the Euclidean sense. E.g., a lightlike vector has a zero inner product with itself. This is because the metric isn't the Euclidean metric. However, the equation $t^a \nabla_a v^b = 0$ doesn't actually involve the metric in the following sense. As a simple example, say you have some vector field $\textbf{v}$, spacetime is flat, you're using coordinates $(p,q)$, and you want to know whether a piece of the field lying along the $p$ axis constitutes parallel transport. Since spacetime is flat in my example, the operator $\nabla_a$ is simply a synonym for $\frac{\partial}{\partial x^a}$. The tangent vector has $t^p=1$ and $t^q=0$. So in this example, the condition for parallel transport is simply
$$ t^p \frac{\partial \textbf{v}}{\partial x^p}=1\cdot\frac{\partial \textbf{v}}{\partial x^p}=0.$$
There is no need for any raising or lowering here. You're simply taking the components of the vector $\textbf{v}$ and differentiating them with respect to the coordinate $p$ (a.k.a. $x^p$). Since the metric isn't involved this really is just a directional derivative, as DanielSank says in his comment.
Of course if spacetime weren't flat, the metric would matter because it would come in to the definition of the covariant derivative, but I don't think this affects the interpretation, which is simply that this is a directional derivative.
